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Homework Help: Displacement Vectors

  1. Sep 2, 2014 #1
    1. The problem statement, all variables and given/known data

    An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the north imparts a constant acceleration of 0.300 m/s2. If the acceleration from the wind lasts for 3.50 s, find the magnitude, r, and direction, θ, of the bird\'s displacement during this time period.

    2. Relevant equations

    Δv = a * Δt
    Magnitude of a 2 component vector: A = √(Ax2 + Ay2)

    3. The attempt at a solution
    4.0 mph = 1.78816 m/s

    I drew a picture, but I'm not even sure I drew it correctly - I've been working on this one problem for about 3 hours now with no correct solution.

    Trying to deconstruct the vectors:

    horizontally (east)
    The bird is flying 1.7886 m/s

    Δv = a * Δt
    Δv = .3m/s^2 * 3.5s = 1.05m/s

    From here I think I can find the angle of displacement, or
    arcsin(1.05m/s) / (1.78816m/s) = 35.95826°

    Then I try to use the formula they gave me, but this is probably where I start to go wrong if not aleady:

    to find magnitude of velocity vector: √(1.78816^2 + 1.05^2) = 2.0734 m/s

    R = (2.0736*3.5s) + 1/2(.3)*3.52

    Using this and several variations, I've found the following (wrong) answers for R (magnitude of displacement): 7.2577, 4.9667 and 2.073

    and this is as far as I can get in the problem. I've watched several videos on the subject and just can't seem to put the problem together properly.

    Thanks in advance for the help!

  2. jcsd
  3. Sep 2, 2014 #2


    User Avatar

    Staff: Mentor

    You need to find the displacement vector for the time period of the wind acceleration, not the change in velocity in the easterly direction.

    So instead of this:

    Δv = a * Δt
    Δv = .3m/s^2 * 3.5s = 1.05m/s

    figure out what the total displacement in the easterly direction is. Then the displacement vector is just the 2 components: the constant displacement in the easterly direction, and the total displacement in the southerly direction.
  4. Sep 2, 2014 #3
    Thanks! That pointed me in the right direction and helped a few things click. I used: x = x0 + vt to get:

    1.7786m/s * 3.5s = 6.2251m (displacement east)

    then used:
    Δv= a * Δt = 1.05m/s (wind velocity to the south)

    magnitude of displacement = Δv * Δt = 1.05 m/s * 6.251s = 6.5363m

    This answer worked, but after answering that last part correctly, in the second part of the problem it reads: The displacement vector is:

    [itex]\vec{r}[/itex] = (6.26m) [itex]\hat{i}[/itex] + (1.84m) [itex]\hat{j}[/itex]

    None of that makes any sense to me, since I haven't seen those numbers before. Is there another way to do the same problem? How did they get those numbers?
  5. Sep 2, 2014 #4
    Nevermind, I found my answer. Case closed, thanks!
  6. Sep 2, 2014 #5


    User Avatar

    Staff: Mentor

    Sweet! :smile:
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